You have found the following ages (in years) of all 6 tigers at your local zoo: $ 20,\enspace 15,\enspace 14,\enspace 5,\enspace 3,\enspace 23$ What is the average age of the tigers at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 6 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{20 + 15 + 14 + 5 + 3 + 23}{{6}} = {13.3\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $20$ years $6.7$ years $44.89$ years $^2$ $15$ years $1.7$ years $2.89$ years $^2$ $14$ years $0.7$ years $0.49$ years $^2$ $5$ years $-8.3$ years $68.89$ years $^2$ $3$ years $-10.3$ years $106.09$ years $^2$ $23$ years $9.7$ years $94.09$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{44.89} + {2.89} + {0.49} + {68.89} + {106.09} + {94.09}} {{6}} $ $ {\sigma^2} = \dfrac{{317.34}}{{6}} = {52.89\text{ years}^2} $ The average tiger at the zoo is 13.3 years old. The population variance is 52.89 years $^2$.